How do you solve Subrectangle Queries on LeetCode?

Subrectangle Queries (LeetCode #1476) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the difference between direct updates and lazy updates can significantly affect performance.

What is the brute force solution for Subrectangle Queries?

The brute force approach for Subrectangle Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach directly modifies the subrectangle by iterating through each element in the specified range and updating it. This is straightforward but can be slow for large rectangles. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subrectangle Queries?

Subrectangle Queries uses the following concepts: Array, Design, Matrix. The recommended patterns to study are: Lazy Propagation, Array Manipulation.

What are the interview tips for Subrectangle Queries?

Always clarify the constraints and expected input size before diving into the solution. Discuss the trade-offs of different approaches with your interviewer. Practice explaining your thought process clearly and concisely.

What are common mistakes when solving Subrectangle Queries?

Not considering the implications of multiple updates on the same subrectangle. Failing to account for the order of updates when retrieving values.

#1476

Subrectangle Queries

Medium

ArrayDesignMatrixLazy PropagationArray Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of updating the rectangle directly, we can keep track of updates using a lazy propagation technique. This allows us to apply updates only when necessary, making it efficient.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array to store the rectangle and an additional array to track updates.
2Step 2: For updateSubrectangle, instead of updating the rectangle directly, store the update information in a separate structure.
3Step 3: For getValue, check if there's an update for the specific cell and return the appropriate value.

solution.py13 lines

1class SubrectangleQueries:
2    def __init__(self, rectangle):
3        self.rectangle = rectangle
4        self.updates = []
5
6    def updateSubrectangle(self, row1, col1, row2, col2, newValue):
7        self.updates.append((row1, col1, row2, col2, newValue))
8
9    def getValue(self, row, col):
10        for (row1, col1, row2, col2, newValue) in reversed(self.updates):
11            if row1 <= row <= row2 and col1 <= col <= col2:
12                return newValue
13        return self.rectangle[row][col]

ℹ

Complexity note: The time complexity for getValue is O(n) in the worst case, where n is the number of updates. The space complexity is O(n) due to storing updates.

1Understanding the difference between direct updates and lazy updates can significantly affect performance.
2Tracking updates separately allows for more efficient retrieval of values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.